Jason Yuhas

2021-12-31

Solve: $3\mathrm{sin}2x+4\mathrm{cos}2x-2\mathrm{cos}x+6\mathrm{sin}x-6=0$
My Try
$6\mathrm{sin}x\mathrm{cos}x+4\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)-2\mathrm{cos}x+6\mathrm{sin}x-6=0$

Timothy Wolff

Expert

$6\mathrm{sin}x\mathrm{cos}x+4\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)-2\mathrm{cos}x+6\mathrm{sin}x-6=0$
$6\mathrm{sin}x\mathrm{cos}x-9{\mathrm{sin}}^{2}x-{\mathrm{cos}}^{2}x-2\mathrm{cos}x+6\mathrm{sin}x-1=0$
$-{\left(3\mathrm{sin}x-\mathrm{cos}x\right)}^{2}+2\left(3\mathrm{sin}x-\mathrm{cos}x\right)-1=0$
${\left(3\mathrm{sin}x-\mathrm{cos}x\right)}^{2}-2\left(3\mathrm{sin}x-\mathrm{cos}x\right)+1=0$

Philip Williams

Expert

x=2y we get after some manipulation
$-4{\mathrm{cos}}^{4}y-36{\mathrm{cos}}^{2}y{\mathrm{sin}}^{2}y+24{\mathrm{cos}}^{3}y\mathrm{sin}y=0$
$-4{\mathrm{cos}}^{2}y{\left(\mathrm{cos}y-3\mathrm{sin}y\right)}^{2}=0$

Vasquez

Expert

Abbreviating $\mathrm{sin}x$ and $\mathrm{cos}x$ to s and c, and writing $\mathrm{sin}2x=2sc$ and $\mathrm{cos}2x=2{c}^{2}-1$, the equation becomes
$6sc+4\left(2{c}^{2}-1\right)-2c+6s-6=0$
Factoring out a 2 and grouping terms, we rewrite this as
$3s\left(c+1\right)+4{c}^{2}-c-5=0$
But $4{c}^{2}-c-5=\left(4c-5\right)\left(c+1\right)$, so we have two solutions in s and c:
c=-1 and 3s+4c=5
Reverting to $\mathrm{sin}x$ and $\mathrm{cos}x$ and recognizing the Pythagorean triple ${3}^{2}+{4}^{2}={5}^{2}$, these become
$\mathrm{cos}x=-1$ and $\mathrm{sin}\left(x+\theta \right)=1$
where $\mathrm{sin}\theta =\frac{4}{5}$ and $\mathrm{cos}\theta =\frac{3}{5}$, e.g. $\theta =\mathrm{arcsin}\left(\frac{4}{5}\right)$. So the solution set is
$\left\{\pi +2n\pi |n\in \mathbb{Z}\right\}\cup \left\{\frac{\pi }{2}-\mathrm{arcsin}\left(\frac{4}{5}\right)+2n\pi |n\in \mathbb{Z}\right\}$

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